121x^2+49=105

Solution for 121x^2+49=105 equation:

121x^2+49=105

We move all terms to the left:

121x^2+49-(105)=0

We add all the numbers together, and all the variables

121x^2-56=0

a = 121; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·121·(-56)
Δ = 27104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{27104}=\sqrt{1936*14}=\sqrt{1936}*\sqrt{14}=44\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-44\sqrt{14}}{2*121}=\frac{0-44\sqrt{14}}{242}
=-\frac{44\sqrt{14}}{242}
=-\frac{2\sqrt{14}}{11}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+44\sqrt{14}}{2*121}=\frac{0+44\sqrt{14}}{242}
=\frac{44\sqrt{14}}{242}
=\frac{2\sqrt{14}}{11}
$